MATHEMATICS SPECIALIST. Calculator-free. ATAR course examination Marking Key
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1 MATHEMATICS SPECIALIST Calculator-free ATAR course examination 8 Marking Key Marking keys are an explicit statement about what the examining panel expect of candidates when they respond to particular examination items. They help ensure a consistent interpretation of the criteria that guide the awarding of marks. 8/675 Copyright School Curriculum and Standards Authority 8
2 MATHEMATICS SPECIALIST CALCULATOR-FREE Section One: Calculator-free Question 5% (5 Marks) (5 marks) Functions f, g are defined such that: f ( x) x ( ) g x x x (a) Determine g f ( x). ( mark) ( ) ( ) g f x g x forms a correct expression for ( ) x x g f x (b) Determine the domain for g f ( x). ( marks) We require x so the square root operation is defined. i.e. x However, we require that x so that division by zero does not occur. i.e. x i.e. x 7 Hence D { x x, x 7}. gof states that x states that x 7 (c) Given that ( ) Explain. No, this is FALSE. f f x x +, is it true that f ( ) 4? ( marks) The domain for ( x) Hence ( ) f is not defined. is x since D Rf. states that the statement is false f is not defined states that ( ) f
3 CALCULATOR-FREE MATHEMATICS SPECIALIST Question (6 marks) (a) Solve the following system of equations: ( marks) ( ) ( ) ( ) 4x y z 5... x+ y z 4... x y z... ( ) ( ) : y ( ) ( ) y 4: y+ z 7 ( ) + z 7 z 5 ( ) x ( ) ( ) : x x Hence the solution is unique: x, y, z 5. uses appropriate alegbra correctly with two pairs of equations solves correctly to find the first variable solves correctly to find the second and third variables Alternative 4 5 R R R R R 4R 7 y i.e. y y+ z 7 i.e. ( ) + z 7 i.e. z 5 4x 5 5 ( ) ( ) i.e. 4x 4 i.e. x Hence the solution is unique: x, y, z 5. applies at least two correct row operations solves correctly to find the first variable solves correctly to find the second and third variables
4 MATHEMATICS SPECIALIST 4 CALCULATOR-FREE Question (continued) Consider another set of equations where k is a constant. x y z x y z x y+ kz 6 It can be shown that this system of equations can be reduced to the following: x y z ( k ) ( k + ) ( k ) ( k + ) k + (b) Explain whether this system of equations will have a unique solution for all real values of k. If not, explain the geometric interpretation of this. ( marks) This system of equations will have a unique solution for all values of k provided k. When k there will be NO solution. This is due to there being TWO planes that are PARALLEL to each other (equations and ). states there is a unique solution for k states that k will yield no solution states the geometric interpretation for k i.e. two planes are parallel
5 CALCULATOR-FREE 5 MATHEMATICS SPECIALIST Question (9 marks) (a) Let z a + bi be any complex number. ( marks) z i Obtain an equation relating ab, given that Re. z z i a+ ( b ) i a+ ( b ) i a bi... ( ) z a + bi a + bi a bi a + b( b ) ai... ( ) a + b Hence if the real part is ZERO then it must be true that: a+ bb ( ). i.e. a + b b forms the correct expression equivalent to ( ) forms the correct expression equivalent to ( ) forms the equation relating ab, stating the real part is zero (b) Let z rcisθ be any complex number. Obtain an expression for: (i) i z in terms of r, θ. ( marks) cis i cis θ + z rcis ( θ ) r converts i into polar form correctly writes the correct expression for z in polar form divides polar forms correctly in terms of r, θ
6 MATHEMATICS SPECIALIST 6 CALCULATOR-FREE Question (continued) (ii) arg ( z r) + in terms of θ. ( marks) Let z OP rcisθ and r PQ. OPQR is a rhombus with side length r. Then the complex number z + r OQ is a diagonal in the rhombus. It is a property that a diagonal bisects the angles in a rhombus. s QOR θ s POR i.e. arg ( z+ r) indicates the vector position for z+ r correctly identifies z+ r as the diagonal of a rhombus arg z + r writes the correct expression for ( )
7 CALCULATOR-FREE 7 MATHEMATICS SPECIALIST Question 4 (4 marks) The graph of f ( x) ( )( ) ( x c)( x d) k x a x b is shown below. Determine the value of the constants a, b, c, d and k. a b c d k - - Explain your choice for the value of k. Horizontal intercepts are x, x a, b Vertical asymptotes are x, x c, d Horizontal asymptote is y k states the values for a, b correctly states the values for c, d correctly states the values for k correctly explains/justifies the value for k
8 MATHEMATICS SPECIALIST 8 CALCULATOR-FREE Question 5 (4 marks) Using the substitution u cos( x) 4 8 cos sin ( x ) ( ) x dx., evaluate exactly the definite integral Put u cos( x) du du sin ( x) i. e. dx dx sin When x, u and x, u u cos ( x) sin ( x) dx du 9 uses the substitution ( ) changes the limits correctly anti-differentiates correctly evaluates correctly ( x) u u cos x to express the integrand correctly in terms of u OR Alternative ( ) ( ) ( ) ( ) cos x sin x dx sin x cos x dx 4 d dx ( cos( ))( cos( )) 9 4 ( x) 8 x x dx cos cos cos ( ) 8 [ ] 8 8 identifies sin ( x ) as part of the derivative of cos( x ) uses the factor to express the definite integral anti-differentiates correctly using the next highest power evaluates correctly
9 CALCULATOR-FREE 9 MATHEMATICS SPECIALIST Question 6 (5 marks) (a) Given that ( )( ) a b +, determine the values for a and b. ( marks) x+ x x x+ a b a( x+ ) + b( x ) a+ b x + a b + x x+ x+ x x+ x ( )( ) ( ) ( ) ( )( ) Hence equating co-efficients we obtain a+ b a b i.e. a, b. obtains the numerator correctly in simplifying the algebraic fractions determines the values for a and b correctly (b) Hence determine x dx. ( marks) dx dx + ( )( ) x x x dx x x+ ln ln x x+ + c x ln x + + c writes the integral in terms of the partial fractions correctly anti-differentiates correctly uses an integration constant
10 MATHEMATICS SPECIALIST CALCULATOR-FREE Question 7 (6 marks) (a) Solve the equation z + giving solutions in polar form rcisθ. ( marks) z cis ( ) z cis k where,, z cis, ( ) 5 z cis, z cis cis i.e. z cis, cis ( ), cis expresses z as cis ( ) states the first root in correct polar form for k states the other roots correctly using the argument separation P z z z + 5z + z z+ 5 can be written in the form It can be shown that ( ) 5 4 ( ) ( ) ( ) P z z + Q z. (b) Determine Q( z ). ( ) 5 4 P z z z z z z z ( z z 5) ( z z 5) ( z )( z z 5) Q z z z+ 5. Hence ( ) determines Q( z ) correctly ( mark)
11 CALCULATOR-FREE MATHEMATICS SPECIALIST (c) 5 4 Hence solve the equation z z + 5z + z z+ 5 giving all solutions in Cartesian form a + bi. ( marks) z z+ 5 z + i.e. Solve z z+ 5 or ( )( ) i.e. ( z ) + 4 or i.e. ( ) ( ) z z + z + i z i or z, cis, cis i.e. z +, i, i, + i, i i.e. z ± i,, ± i states z ± i as solutions states z ± i as solutions
12 MATHEMATICS SPECIALIST CALCULATOR-FREE Question 8 (5 marks) A parallelepiped is a prism where each face is a parallelogram. Let OAPB be the parallelogram formed by the horizontal sides a OA and b OB where a 6 and b 8. The third side that forms the parallelepiped is c OC where c. 5 Let θ the size of AOB φ the angle between OC and the positive z axis (a) Determine a b. 8 6 ( ) ( ) a b 6 ( 8) ( ) ( ) 6( 8) 54 uses the correct form for each component for the cross product evaluates each component correctly ( marks)
13 CALCULATOR-FREE MATHEMATICS SPECIALIST The volume of any prism can be found by considering the formula Volume Area ( Base) h, where h the perpendicular height of the prism. It is also true that a b absinθ. (b) Explain why c ( a b) From Volume Area ( Base) h will determine the volume of the parallelepiped. ( marks) Area(OAPB ) ( OA)( OB) sin a b θ Perpendicular height h ( OC)cosφ Note that a b is parallel to the z-axis hence the angle between c and a b is also φ. Hence volume V c cosφ a b c a b cosφ. i.e. a dot product c ( a b). Since the value cosφ may be less than zero we consider the absolute value of this dot product. i.e. V c ( a b) justifies that a b determines the area of the parallelogram base justifies that the perpendicular height h c cosφ (c) Hence determine the exact volume of the parallelepiped. ( mark) Using V c ( a b) ( ) + ( ) + 5( 54) 7 cubic units 5 54 evaluates the dot product correctly
14 MATHEMATICS SPECIALIST 4 CALCULATOR-FREE Question 9 (7 marks) (a) By using an appropriate trigonometric identity, simplify in terms of u, the expression x x 4 x tan u +. ( marks) + where ( ) x x+ 4 tan u+ tan u+ + 4 ( ) ( ) u+ u+ u + tan u + tan + tan tan tan 4 ( u ) sec u expands correctly to obtain tan u + uses the trigonometric identity correctly to simplify to sec u (b) Hence evaluate dx exactly. (5 marks) ( x x+ 4) dx Using x tan ( u) + sec u i. e. dx sec u. du du When x, u and x, u dx sec u. du sec u. du du secu sec u ( x x+ 4) ( sec u) 6 cos u du sin u 6 ( ) 6 obtains dx correctly in terms of du changes the limits correctly in terms of u simplifies the integrand correctly in terms of u using the expression from part (a) anti-differentiates correctly evaluates correctly
15 This document apart from any third party copyright material contained in it may be freely copied, or communicated on an intranet, for non-commercial purposes in educational institutions, provided that it is not changed and that the School Curriculum and Standards Authority is acknowledged as the copyright owner, and that the Authority s moral rights are not infringed. Copying or communication for any other purpose can be done only within the terms of the Copyright Act 968 or with prior written permission of the School Curriculum and Standards Authority. Copying or communication of any third party copyright material can be done only within the terms of the Copyright Act 968 or with permission of the copyright owners. Any content in this document that has been derived from the Australian Curriculum may be used under the terms of the Creative Commons Attribution 4. International (CC BY) licence. Published by the School Curriculum and Standards Authority of Western Australia Sevenoaks Street CANNINGTON WA 67
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